JEE MAIN - Physics (2018 (Offline) - No. 5)
The dipole moment of a circular loop carrying a current I, is m and the magnetic field at the centre of the
loop is B1. When the dipole moment is doubled by keeping the current constant, the magnetic field at the
centre of the loop is $${{B_2}}$$. The ratio $${{{B_1}} \over {{B_2}}}$$ is:
2
$$\sqrt 3 $$
$$\sqrt 2 $$
$$1 \over \sqrt 2 $$
Explanation
Dipole moment, M = IA
Let radius of circular loop = R
$$\therefore\,\,\,$$ M = I $$ \times $$ $$\pi $$R2
Later, we keep current constant ,
But dipole moment becomes double, let new radius = R1
$$\therefore\,\,\,$$ 2M = I $$ \times $$ $$\pi $$R$$_1^2$$
$$ \Rightarrow $$$$\,\,\,$$ 2I$$\pi $$R2 = I$$\pi $$R$$_1^2$$
$$ \Rightarrow $$$$\,\,\,$$ R1 = $$\sqrt 2 $$ R
At the center of circular ring, the magnetic field,
B = $${{{\mu _0}I} \over {2R}}$$
$$\therefore\,\,\,$$ B1 = $${{{\mu _0}I} \over {2R}}$$ and B2 = $${{{\mu _0}I} \over {2 \times \left( {\sqrt 2 R} \right)}}$$
$$\therefore\,\,\,$$ $${{{B_1}} \over {{B_2}}}$$ = $${{{{{\mu _0}I} \over {2R}}} \over {{{{\mu _0}I} \over {2\sqrt 2 \,R}}}}$$
= $${{2\sqrt 2 } \over 2}$$
= $$ \sqrt 2 $$
Let radius of circular loop = R
$$\therefore\,\,\,$$ M = I $$ \times $$ $$\pi $$R2
Later, we keep current constant ,
But dipole moment becomes double, let new radius = R1
$$\therefore\,\,\,$$ 2M = I $$ \times $$ $$\pi $$R$$_1^2$$
$$ \Rightarrow $$$$\,\,\,$$ 2I$$\pi $$R2 = I$$\pi $$R$$_1^2$$
$$ \Rightarrow $$$$\,\,\,$$ R1 = $$\sqrt 2 $$ R
At the center of circular ring, the magnetic field,
B = $${{{\mu _0}I} \over {2R}}$$
$$\therefore\,\,\,$$ B1 = $${{{\mu _0}I} \over {2R}}$$ and B2 = $${{{\mu _0}I} \over {2 \times \left( {\sqrt 2 R} \right)}}$$
$$\therefore\,\,\,$$ $${{{B_1}} \over {{B_2}}}$$ = $${{{{{\mu _0}I} \over {2R}}} \over {{{{\mu _0}I} \over {2\sqrt 2 \,R}}}}$$
= $${{2\sqrt 2 } \over 2}$$
= $$ \sqrt 2 $$
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